Recent questions in Advanced Math

Upper Level MathAnswered question

Messiah Sutton 2022-11-09

The velocity distribution for laminar flow between parallel plates is given by:

$\frac{u}{{u}_{max}}=1-{\left(\frac{2y}{h}\right)}^{2}$

where h is the distance separating the plates and the origin is placed midway between the plates. Consider a flow of water at ${15}^{\circ}C$ , with ${u}_{max}=0.30\frac{m}{sec}$ and $h=0.50\text{}mm$ . Calculate the shear stress on the upper plate and give its direction.

$\frac{u}{{u}_{max}}=1-{\left(\frac{2y}{h}\right)}^{2}$

where h is the distance separating the plates and the origin is placed midway between the plates. Consider a flow of water at ${15}^{\circ}C$ , with ${u}_{max}=0.30\frac{m}{sec}$ and $h=0.50\text{}mm$ . Calculate the shear stress on the upper plate and give its direction.

Discrete mathAnswered question

Jared Lowe 2022-11-05

How to arrive at this integral?

However, I am not sure how to arrive at this:

$$\begin{array}{rl}\phantom{=}& {{\int}_{0}^{1}{x}^{n+2}(1-x{)}^{n}\phantom{\rule{thinmathspace}{0ex}}dx}+2{\int}_{0}^{1}{x}^{n+1}(1-x{)}^{n+1}\phantom{\rule{thinmathspace}{0ex}}dx+{{\int}_{0}^{1}{x}^{n}(1-x{)}^{n+2}\phantom{\rule{thinmathspace}{0ex}}dx}\\ =& 2I(n+1)+{2{\int}_{0}^{1}{x}^{n+2}(1-x{)}^{n}\phantom{\rule{thinmathspace}{0ex}}dx}\end{array}$$

and why the highlighted part is equal to each other? What type of integral is that?

However, I am not sure how to arrive at this:

$$\begin{array}{rl}\phantom{=}& {{\int}_{0}^{1}{x}^{n+2}(1-x{)}^{n}\phantom{\rule{thinmathspace}{0ex}}dx}+2{\int}_{0}^{1}{x}^{n+1}(1-x{)}^{n+1}\phantom{\rule{thinmathspace}{0ex}}dx+{{\int}_{0}^{1}{x}^{n}(1-x{)}^{n+2}\phantom{\rule{thinmathspace}{0ex}}dx}\\ =& 2I(n+1)+{2{\int}_{0}^{1}{x}^{n+2}(1-x{)}^{n}\phantom{\rule{thinmathspace}{0ex}}dx}\end{array}$$

and why the highlighted part is equal to each other? What type of integral is that?

Discrete mathAnswered question

MMDCCC50m 2022-11-05

Find a recursion formula for combinatorial problem

Let ${C}_{n}$ be the number of sequences with a length of n, which their elements belong to {0,1,2}, and they don't contain the following sequences: 11,21. Find a recursion formula with starting conditions for ${C}_{n}$.

Let ${C}_{n}$ be the number of sequences with a length of n, which their elements belong to {0,1,2}, and they don't contain the following sequences: 11,21. Find a recursion formula with starting conditions for ${C}_{n}$.

Discrete mathAnswered question

Noe Cowan 2022-11-02

What is the most general distribution for which $E[1/x]=1/E[x]$?

What is the most general distribution for which the expected value of the multiplicative inverse equals the multiplicative inverse of the expected value?

Motivation: I'm into modelling dynamics on graphs and I found a problem which is easily solvable in cases where the degree distribution of the vertices is a distribution where $E[1/k]=1/E[k]$. (${k}_{i}$ is the degree of the ith vertex) From this solution I may gain an insight into how to unify multiple models.

So particularly I'm looking for a distribution which consists of non-negative, finite integers. But I'm also interested in continuous solutions. Distributions where $E[1/{k}^{n}]=1/E[{k}^{n}]$ may also help unifying the models.

What I do know so far, that ${k}_{i}=1$ is a particular solution. In the continuous case every function where $f(x)=f(1/x)$ and $E[x]=1$ is a solution. I know what momentum generating functions are and they seem like a good direction to try in, but I failed so far.

What is the most general form of this distribution? Does it have a name? It sounds like something trivial, like a "famous" distribution, but I can't find it.

What is the most general distribution for which the expected value of the multiplicative inverse equals the multiplicative inverse of the expected value?

Motivation: I'm into modelling dynamics on graphs and I found a problem which is easily solvable in cases where the degree distribution of the vertices is a distribution where $E[1/k]=1/E[k]$. (${k}_{i}$ is the degree of the ith vertex) From this solution I may gain an insight into how to unify multiple models.

So particularly I'm looking for a distribution which consists of non-negative, finite integers. But I'm also interested in continuous solutions. Distributions where $E[1/{k}^{n}]=1/E[{k}^{n}]$ may also help unifying the models.

What I do know so far, that ${k}_{i}=1$ is a particular solution. In the continuous case every function where $f(x)=f(1/x)$ and $E[x]=1$ is a solution. I know what momentum generating functions are and they seem like a good direction to try in, but I failed so far.

What is the most general form of this distribution? Does it have a name? It sounds like something trivial, like a "famous" distribution, but I can't find it.

Discrete mathAnswered question

Joglxym 2022-11-02

Let $f:A\to B$ and $g:C\to D$. Define$$f\times g=\{((a,c),(b,d)):(a,b)\in f\text{and}(c,d)\in g\}.$$Prove that $f\times g$ is a function from $A\times C$ to $B\times D$.

My proof (so far):

Let $(a,c)\in A\times C$, then $a\in A$ and $c\in C$.

Let $(b,d)\in B\times D$, $b\in B$ and $d\in D$.

If $p\in A\times C$ then $p=(a,c)$.

Define $(f\times g)(p)$ as $(f\times g)(p)=(f\times g)(a,c)=(f(a),g(c))=(b,d)$, however $b\in B$ and $d\in D$ and $(b,d)\in B\times D$.

This shows $f\times g$ is a function from $A\times C$ to $B\times D$ as $(f\times g):A\times C\to B\times D$ then $(a,b)\to (f(a),g(c))$.

I know that it is not well written out, but I was wondering if my thought process so far made sense or if I accidentally missed something. Additionally, I am unsure of what the next step may be. As a sidenote, I am worried that this does not work for the given definition of $f\times g$.

My proof (so far):

Let $(a,c)\in A\times C$, then $a\in A$ and $c\in C$.

Let $(b,d)\in B\times D$, $b\in B$ and $d\in D$.

If $p\in A\times C$ then $p=(a,c)$.

Define $(f\times g)(p)$ as $(f\times g)(p)=(f\times g)(a,c)=(f(a),g(c))=(b,d)$, however $b\in B$ and $d\in D$ and $(b,d)\in B\times D$.

This shows $f\times g$ is a function from $A\times C$ to $B\times D$ as $(f\times g):A\times C\to B\times D$ then $(a,b)\to (f(a),g(c))$.

I know that it is not well written out, but I was wondering if my thought process so far made sense or if I accidentally missed something. Additionally, I am unsure of what the next step may be. As a sidenote, I am worried that this does not work for the given definition of $f\times g$.

Upper Level MathOpen question

Benu Sharma2022-10-28

In August 2013, E*TRADE Financial was offering only 0.05% interest on its online checking accounts, with interest reinvested monthly. Find the associated exponential model for the value of a $5000 deposit after "t" years. Assuming that this rate of return continued for 7 years, how much would a deposit of $5000 in August 2013 be worth in August 2020? (Answer to the nearest $1. )

Discrete mathAnswered question

Maribel Vang 2022-10-28

Extracting even / odd part of summation trick

Given a function f(x), we know its even part is given as $\frac{f(x)+f(-x)}{2}$ and its odd part is given by $\frac{f(x)-f(-x)}{2}$.

Consider a discrete sequence given by ${a}_{j}$ for $j\ge 1$, then the sum of the terms in the sequence till n terms is given as

$$S=\sum _{j}^{n}{a}_{j}$$

Suppose I wanted to get the sum of the even terms in the above expression; then

$${S}_{odd}=\sum _{j}^{n}\frac{{a}_{-j}+{a}_{j}}{2}$$

and, for odd,

$${S}_{odd}=\sum _{j}^{n}\frac{{a}_{j}-{a}_{-j}}{2}$$

But wait, our sequence was defined for $j\ge 1$. Well, here's the thing: ${a}_{j}$ is some function of j, extending the domain to negative integers and evaluating the function gives the right answer... but I can't understand why the continuous function trick extended to here.

Examples:

Sum of first n numbers given as $\frac{n(n+1)}{2}$, sum of first n odds will be given as: $\frac{n(n+1)}{2}-\frac{n(1-n)}{2}$

My attempt at finding an exact connection: To the discrete sequence $\frac{n(n+1)}{2}$, we can associate a function $f(x)=\frac{x(x+1)}{2}$ and we can think of the summation as summing this function at several different input points i.e:

$$S=\sum _{j}^{n}{a}_{j}\to \sum _{k}^{n}f(x+k)$$

Then we apply the even odd decomposition and return back to the sequence world.

My question: Does there exist an association for every sequence with a function? If not, what is the criterion for an association to exist?

Given a function f(x), we know its even part is given as $\frac{f(x)+f(-x)}{2}$ and its odd part is given by $\frac{f(x)-f(-x)}{2}$.

Consider a discrete sequence given by ${a}_{j}$ for $j\ge 1$, then the sum of the terms in the sequence till n terms is given as

$$S=\sum _{j}^{n}{a}_{j}$$

Suppose I wanted to get the sum of the even terms in the above expression; then

$${S}_{odd}=\sum _{j}^{n}\frac{{a}_{-j}+{a}_{j}}{2}$$

and, for odd,

$${S}_{odd}=\sum _{j}^{n}\frac{{a}_{j}-{a}_{-j}}{2}$$

But wait, our sequence was defined for $j\ge 1$. Well, here's the thing: ${a}_{j}$ is some function of j, extending the domain to negative integers and evaluating the function gives the right answer... but I can't understand why the continuous function trick extended to here.

Examples:

Sum of first n numbers given as $\frac{n(n+1)}{2}$, sum of first n odds will be given as: $\frac{n(n+1)}{2}-\frac{n(1-n)}{2}$

My attempt at finding an exact connection: To the discrete sequence $\frac{n(n+1)}{2}$, we can associate a function $f(x)=\frac{x(x+1)}{2}$ and we can think of the summation as summing this function at several different input points i.e:

$$S=\sum _{j}^{n}{a}_{j}\to \sum _{k}^{n}f(x+k)$$

Then we apply the even odd decomposition and return back to the sequence world.

My question: Does there exist an association for every sequence with a function? If not, what is the criterion for an association to exist?

Upper Level MathOpen question

Anyangwa Emmanuel Atekwana2022-10-16

Write an equivalent first-order differential equation and initial condition for y

y=−1+∫x1(t−y(t))dt

Upper Level MathOpen question

Elizabet Rodriguez2022-09-28

**a. **Write the formula for the volume V of a prism where B is the area of the base and h$h$ is the height.

V=

**b. **Solve the formula for B$B$.

B=

**c. **Use the new formula to find the area of the base of the prism.

The area of the base of the prism is square inches.

Students pursuing advanced Math are constantly dealing with advanced Math equations that are mostly used in space engineering, programming, and construction of AI-based solutions that we can see daily as we are turning to automation that helps us to find the answers to our challenges. If it sounds overly complex with subjects like exponential growth and decay, don’t let advanced math problems frighten you because these must be approached through the lens of advanced Math questions and answers. Regardless if you are dealing with simple equations or more complex ones, just break things down into several chunks as it will help you to find the answers.